Date | May 2013 | Marks available | 7 | Reference code | 13M.2.sl.TZ1.7 |
Level | SL only | Paper | 2 | Time zone | TZ1 |
Command term | Find | Question number | 7 | Adapted from | N/A |
Question
A random variable \(X\) is normally distributed with \(\mu = 150\) and \(\sigma = 10\) .
Find the interquartile range of \(X\) .
Markscheme
recognizing one quartile probability (may be seen in a sketch) (M1)
eg \({\rm{P}}(X < {Q_3}) = 0.75\) , \(0.25\)
finding standardized value for either quartile (A1)
eg \(z = 0.67448 \ldots \) , \(z = - 0.67448 \ldots \)
attempt to set up equation (must be with \(z\)-values) (M1)
eg \(0.67 = \frac{{{Q_3} - 150}}{{10}}\) , \( - 0.67448 = \frac{{x - 150}}{{10}}\)
one correct quartile
eg \({Q_3} = 156.74 \ldots \) , \({Q_1} = 143.25 \ldots \)
correct working (A1)
eg other correct quartile, \({Q_3} - \mu = 6.744 \ldots \)
valid approach for IQR (seen anywhere) (A1)
eg \({Q_3} - {Q_1}\) , \(2({Q_3} - \mu )\)
IQR \( = 13.5\) A1 N4
[7 marks]
Examiners report
This was an accessible problem that created difficulties for candidates. Although they recognized and often wrote down a formula for IQR, most did not understand the conceptual nature of the first and third quartiles. Those who did could solve the problem effectively using their GDC in relatively few steps. Candidates that were able to start this question often drew the normal curve and gave quartile values at \(140\) and \(160\). This generally led to a solution which while wrong, was also clearly inadequate for the indicated 7 marks.